Transcript Level of Service - University of Evansville

Level of Service
Section 4.5
Level of Service: The Basics
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Measure performance by density.
Ratio of operating capacity to speed.
Units: passenger cars per mile per lane
(pc/mi/ln)
Denoted by the letters A-F
These letters represent the Level of
Service (LOS) of the freeway.
Level of Service & Density Ranges
LOS
Density Range (pc/mi/ln)
A
0-11
B
>11-18
C
>18-26
D
>26-35
E
>35-45
F
>45
(Values taken from 2000 Highway Capacity Manual, Exhibit 23-2)
Freeway Base Conditions
2 foot
left side
clearance
5 lanes per
direction
Minimum 6
foot right
side
clearance
12 foot lane
width
Freeway Base Conditions
Access spacing 2 miles or greater
Freeway Base Conditions
•All Passenger Car Composition
•Commuter Driver Population
Level Terrain, grades no
greater than 2 percent
We then take those base conditions
and make adjustments to match the
characteristics of our freeway
segment!!
Level of Service is determined by the
density of traffic in a particular segment,
estimated by the following equation:
Vp
D
S
(P&P pg 153 EQ. 4.5.2)
Where…
D = density in pc/mi/ln
Vp = flow rate in pc/h/ln
S = free flow speed in mi/h
Let’s Start at the Top
Vp = Flow Rate (pc/hr/ln)
V
Vp 
PHF  N  fHV  fDP
(P&P pg 153 EQ. 4.5.3)
Where
• V = hourly volume (veh/hr)
• PHF = peak hour factor
• N = number of lanes per direction (ln)
• fHV = heavy vehicle factor
• fDP = driver population factor
Peak Hour Factor (PHF)
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Represents the variation of traffic
flow within one hour.
Typical freeway range is 0.80 to 0.95
Based on local conditions.
The equation for PHF is:
V
V
PHF  
q Nt (60/t)
(P&P pg 151 EQ. 4.5.1)
Where
• V = hourly volume
• q = flow rate
• Nt = number of vehicles counted
during time period t
Peak Hour Factor Example
GIVEN: Vehicles are counted on a freeway in 5
minute intervals. The highest vehicle count for any
one 5 minute segment was 250 vehicles. The total
number of vehicles counted for the hour was 2600.
REQUIRED: Compute the Peak Hour Factor
SOLUTION:
60 
60 


q  Nt    250   3000veh/h
 t 
 5 
2600
PHF 
 0.87
3000
The peak hour factor for this segment of freeway is 0.87
Note that this value is in the typical range for freeway PHF.
Heavy Vehicle Factor
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Heavy vehicles converted to passenger cars
using the Heavy Vehicle Factor, fHV
fHV
Where
1

1  PT (ET  1)  PR (ER  1)
(P&P pg 154 EQ. 4.5.4)
• PT = proportion of trucks and buses in traffic
• PR = proportion of recreational vehicles in traffic
• ET = passenger-car equivalent for trucks and buses
• ER = passenger car equivalent for recreational vehicles
These equivalent values can be found in HCM
2000, Exhibit 23-8 or P&P pg. 154
Heavy Vehicle Factor Example
GIVEN: From the previous PHF example, 2600 vehicles are
counted on a level freeway segment, of which 90 were
trucks & buses and 5 were recreational vehicles.
REQUIRED: Find the heavy vehicle factor for the hour.
Hint: Pay attention to the type of terrain.
SOLUTION:
Using the equivalent passenger-car values for level terrain:
fHV 
1
90
5
1
(1.5  1) 
(1.2  1)
2600
2600
 0.98
The Heavy Vehicle Factor for this segment is 0.98
Driver Population Factor, fDP
•Changes the flow rate based driver familiarity.
• Ranges from 1.00 (Commuters) to 0.85 (Tourists
and people unfamiliar with the facility).
•Note: a driver population that is most familiar with
the facility does not affect the rate of flow at all.
Flow Rate Example
Now that we have all the parts, let’s calculate the flow rate
of the freeway segment from our previous examples.
GIVEN: 6 lane freeway in an urban area, level terrain, 2600 vehicles
counted in a given hour, and a commuter population.
Recall from previous examples that PHF = 0.87 and fHV = 0.98
REQUIRED: Compute the Flow Rate VP for the segment.
SOLUTION:
Using N = 3 and fDP = 1.00
2600
Vp 
 1016pc/h/l n
(0.87)(3)(0.98)(1.00 )
Free-Flow Speed
The bottom of the LOS equation is the free flow speed
(FFS) of the highway segment. Free flow speed is
measured in miles per hour. The equation for FFS is:
FFS  BFFS  FLW  FLC  FN  FID
(P&P pg 154 EQ. 4.5.5)
Where
•
BFFS = base free flow speed, usually 70 mi/hr (urban) or 75 mi/hr (rural)
•
FLW = Factor for lane width shorter than 12 feet (HCM Exhibit 23-4)
•
FLC = factor for lateral (right side) clearance under 6 feet (HCM Exhibit 23-5)
•
FN = factor for number of lanes less than 5 lanes per direction (HCM Ex 23-6)
•
FID = factor for density of interchanges per mile (HCM Exhibit 23-7)
Free Flow Speed Example
Continuing our previous example, let’s now calculate
the free flow speed for the freeway segment.
GIVEN: Added information to previous example, the
width of the lanes is 11 feet, the right-side clearance is
5 feet, and there are approximately 1.0 interchanges
per mile.
Recall: Urban area, number of lanes N=3
Assume: BFFS = 70 mi/hr
REQUIRED: Calculate the base free-flow speed. Use
values from the HCM 2000.
Free Flow Speed Example Solution
SOLUTION:
FFS  BFFS  FLW  FLC  FN  FID
Lane Width: 11 feet => FLW = 1.9 mi/h
Lateral Clearance: 5 feet, 3 lanes => FLC = 0.4 mi/h
Number of Lanes: 3 => FN = 3.0 mi/h
Interchanges: 1.0 per mile => FID = 2.5 mi/h
FFS  70 1.9  0.4  3.0  2.5  62.2 mi/h
The Free-Flow Speed for this segment, based on the
local conditions, is
62.2 miles/hour.
Density Calculation
Now that we have all the parts, let’s calculate the
density of our freeway segment.
GIVEN: Urban freeway segment.
Vp = 1016 pc/h/ln and S = 62.2 mi/h
REQUIRED: Compute the density and
estimate the level of service.
Density Solution
SOLUTION:
Vp 1016 pc/h/ln
D

 16.33 pc/mi/ln
62.2 mi/h
S
Density Solution
FFS = 70 mi/h
Flow Rate = 1016 pc/h/ln
The density for the segment is 16.33 pc/mi/ln.
Referring to HCM 2000, Exhibit 23-2, this value
corresponds with
Level of Service B
LOS in Action
What does it all mean? Click the picture below to
get a real life view of Level of Service A:
LOS in Action, II
Now click below to see a shot of LOS C in action: